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\begin{document}

\title{An IV random coefficients probit model}
\author{Andrew Chesher \\
%EndAName
CeMMAP \& UCL}
\date{November 29th 2010}
\maketitle

\begin{abstract}
These notes consider the identifying power of a random coefficients probit
model in which a bivariate unobservable is distributed independently of
instruments but may be jointly dependently distributed with the explanatory
variable.
\end{abstract}

\section{Model}

A scalar binary outcome is determined by a structural function $h$, as
follows.%
\begin{equation*}
Y=h(X,U)
\end{equation*}%
In the case considered $X$ is scalar and $U=(U_{1},U_{2})$ is a bivariate
continuously distributed random variable.

The idea is to investigate the identifying power of IV models with high
dimensional unobservables. So while $Y$ is unidimensional, $U$ is two
dimensional. To keep computations relatively straightforward I specify a
quite restrictive model, as follows.

The structural function $h$ is specified as a random coefficients probit
model structural function. It is convenient to have $U\in 
%TCIMACRO{\U{211d} }%
%BeginExpansion
\mathbb{R}
%EndExpansion
^{2}$. Here is the model.%
\begin{equation*}
Y=h(X,U)\equiv \left\{ 
\begin{array}{ccrcl}
0 & , & -\infty < & U_{1} & \leq \alpha _{0}+(\alpha _{1}+U_{2})X \\ 
1 & , & \alpha _{0}+(\alpha _{1}+U_{2})X< & U_{1} & \leq \infty%
\end{array}%
\right.
\end{equation*}%
\begin{equation*}
U%
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\indt%
%EndExpansion
Z
\end{equation*}%
\begin{equation*}
U_{1}\sim N(0,1)\qquad U_{2}|U_{1}=u_{1}\sim N(\beta _{1}u_{1},\beta _{2})
\end{equation*}%
equivalently:%
\begin{equation*}
U\sim N\left( \left( 
\begin{array}{c}
0 \\ 
0%
\end{array}%
\right) ,\left( 
\begin{array}{cc}
1 & \beta _{1} \\ 
\beta _{1} & \beta _{2}+\beta _{1}^{2}%
\end{array}%
\right) \right) .
\end{equation*}%
Let $\mathcal{Z}$ denote the set of values taken by instrumental variables
(which may be scalar or vector, discrete or continuous).

Define $\beta \equiv (\beta _{1},\beta _{2})$. Let $F_{U}(\cdot ,\beta )$
denote the distribution of $U$ and for any $S\subseteq 
%TCIMACRO{\U{211d} }%
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\mathbb{R}
%EndExpansion
^{2}$ let $F_{U}(S,\beta )$ denote the probability that $U\in S$ when $U$
has the bivariate normal distribution specified above with parameter vector $%
\beta .$

\section{Identification}

Define sets $T(0,x;h)$ and $T(1,x;h)$ as follows.%
\begin{eqnarray*}
T(0,x;h) &\equiv &\{u:h(x,u)=0\} \\
T(1,x;h) &\equiv &\{u:h(x,u)=1\}
\end{eqnarray*}%
Here $u=(u_{1},u_{2})$ and because $Y$ is binary, $T(1,x;h)=T(0,x;h)^{C}$.%
\footnote{$A^{C}$ denotes the complement of set $A$.}

The 

sets $T(0,x;h)$ and $T(1,x;h)$ are linear half spaces which partition $%
%TCIMACRO{\U{211d} }%
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\mathbb{R}
%EndExpansion
^{2}$, as follows.%
\begin{eqnarray*}
T(0,x;h) &=&\{u:u_{1}\leq (\alpha _{0}+\alpha _{1}x)+xu_{2}\} \\
T(1,x;h) &\equiv &\{u:u_{1}>(\alpha _{0}+\alpha _{1}x)+xu_{2}\}
\end{eqnarray*}%
Considering any two distinct values $x^{\prime }$ and $x^{\prime \prime }$,
the line separating $T(0,x^{\prime };h)$ and $T(1,x^{\prime };h)$:%
\begin{equation*}
u_{1}=(\alpha _{0}+\alpha _{1}x^{\prime })+x^{\prime }u_{2}
\end{equation*}%
and the line separating $T(0,x^{\prime \prime };h)$ and $T(1,x^{\prime
\prime };h)$:%
\begin{equation*}
u_{1}=(\alpha _{0}+\alpha _{1}x^{\prime \prime })+x^{\prime \prime }u_{2}
\end{equation*}%
intersect at the point $(u_{1},u_{2})=(\alpha _{0},-\alpha _{1})$.

We consider the identified set for $\{h,F_{U}\}$ which in this case is an
identified set for parameters $\theta \equiv (\alpha ,\beta )$ where $\alpha
\equiv (\alpha _{0},\alpha _{1})$.

Consider a set $S\subseteq 
%TCIMACRO{\U{211d} }%
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\mathbb{R}
%EndExpansion
^{2}$ and probability distributions $F_{YX|Z}^{0}$ for $z\in \mathcal{Z}$.
If $\{h,F_{U}\}$ (equivalently $\theta $) is in the identified set then for
all $z\in \mathcal{Z}$:%
\begin{equation}
\Pr [U\in S|X=x,Z=z]\geq \sum_{i\in \{0,1\}}1[T(i,x;h)\subseteq
S]\Pr\nolimits_{0}[Y=i|X=x,Z=z]  \label{in1xz}
\end{equation}%
and on taking expectations over $X$ given $Z=z$%
\begin{equation}
F_{U}(S,\beta )\geq \int \sum_{i\in \{0,1\}}1[T(i,x;h)\subseteq
S]\Pr\nolimits_{0}[Y=i|X=x,Z=z]dF_{X|Z=z}^{0}(x)  \label{in1z}
\end{equation}%
and on maximising over $Z$:%
\begin{equation}
F_{U}(S,\beta )\geq \max_{z\in \mathcal{Z}}\int \sum_{i\in
\{0,1\}}1[T(i,x;h)\subseteq S]\Pr\nolimits_{0}[Y=i|X=x,Z=z]dF_{X|Z=z}^{0}(x).
\label{in1}
\end{equation}

Since $F_{U}$ is parametrically specified, considering all $S\subseteq 
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\mathbb{R}
%EndExpansion
^{2}$ the expression (\ref{in1}) delivers a system of inequalities in $%
\theta $ which define an outer region for the identified set for $\theta $.
In fact this is the identified set. The question to be answered is - is this
set informative?

We know that if $\Pr_{0}[Y=y|Z=z]$ depends on $z$ then it must exclude
values of $\theta $ in which $\alpha _{1}=0$ so it seems certain that it is
informative - so, the question must be, how informative.

We do not need to consider all sets $S\subseteq 
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\mathbb{R}
%EndExpansion
^{2}$. For example it is easy to show that we only need to consider
connected sets.

\section{Discrete $X$}

Consider the case in which $X$ is \emph{discrete} with support $%
\{x_{1},\dots ,x_{K}\}$. Let $\delta _{k}^{0}(z)$ be a shorthand notation
for $\Pr_{0}[X=x_{k}|Z=z]$. Partition $%
%TCIMACRO{\U{211d} }%
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\mathbb{R}
%EndExpansion
^{2}$ into $2^{K}$ sets:%
\begin{equation}
a(i_{1},\dots ,i_{K},h)\equiv \{u:\bigwedge\limits_{k=1}^{K}\left(
h(x_{k},u)=i_{k}\right) \}  \label{adef}
\end{equation}%
where for each $k\in \{1,\dots ,K\}$, $i_{k}$ takes values in $\{0,1\}$.
Note that in general not all of these sets are non-empty. It is important to
understand that the membership of these sets depends on the structural
function $h$ under consideration.

The sets $a(i_{1},\dots ,i_{K},h)$ can be expressed in terms of the sets $%
T(i,x;h)$ as follows.%
\begin{equation*}
a(i_{1},\dots ,i_{K},h)=\bigcap\limits_{k=1}^{K}T(i_{k},x_{k};h)
\end{equation*}

The sets $S$ to be considered are selections from a list of sets defined as
all possible unions of the sets $a(i_{1},\dots ,i_{K},h)$.

\subsection{Binary $X$}

Consider the case in which $K=2$, that is $X$ is binary. Figure \ref%
{aregions} shows the four elemental sets $a(0,0,h)$, $a(0,1,h)$, $a(1,0,h)$
and $a(1,1,h)$ for a particular value of $\alpha $ and a pair of values of $%
x $.\footnote{%
The values employed are $\alpha =(0,1)$, $x_{1}=0.3$, $x_{2}=1.2$.}

Table \ref{unions1} shows a list of the $2^{4}-1=15$ potentially non-empty
unions of the four elemental sets. The sets are denoted $S_{1},\dots ,S_{15}$%
. The column headed $x_{k}$ ($k\in \{1,2\}$) shows the values of $i\in
\{0,1\}$ (if any) delivered by $1[T(i,x;h)\subseteq S_{j}]$ in (\ref{in1xz})
as $i$ varies in $\{0,1\}$.

Table \ref{bounds1} shows the consequent bounds on $F_{U}(S_{j},\beta )$
that that arise on applying (\ref{in1z}). Only $8$ of the $15$ unions yield
informative inequalities.

The identified set for $\theta $ is the set of values of $\theta $ such that
the following $8$ inequalities hold.%
\begin{eqnarray*}
F_{U}(S_{5},\beta ) &\geq &\max_{z\in \mathcal{Z}}\Pr\nolimits_{0}[Y=0\wedge
X=x_{1}|z] \\
F_{U}(S_{6},\beta ) &\geq &\max_{z\in \mathcal{Z}}\Pr\nolimits_{0}[Y=0\wedge
X=x_{2}|z] \\
F_{U}(S_{9},\beta ) &\geq &\max_{z\in \mathcal{Z}}\Pr\nolimits_{0}[Y=1\wedge
X=x_{2}|z] \\
F_{U}(S_{10},\beta ) &\geq &\max_{z\in \mathcal{Z}}\Pr\nolimits_{0}[Y=1%
\wedge X=x_{1}|z]
\end{eqnarray*}%
\begin{equation*}
F_{U}(S_{11},\beta )\geq \max_{z\in \mathcal{Z}}\Pr\nolimits_{0}[Y=0|z]
\end{equation*}%
\begin{eqnarray*}
F_{U}(S_{12},\beta ) &\geq &\max_{z\in \mathcal{Z}}\left(
\Pr\nolimits_{0}[Y=0\wedge X=x_{1}|z]+\Pr\nolimits_{0}[Y=1\wedge
X=x_{2}|z]\right) \\
F_{U}(S_{13},\beta ) &\geq &\max_{z\in \mathcal{Z}}\left(
\Pr\nolimits_{0}[Y=1\wedge X=x_{1}|z]+\Pr\nolimits_{0}[Y=0\wedge
X=x_{2}|z]\right)
\end{eqnarray*}%
\begin{equation*}
F_{U}(S_{14},\beta )\geq \max_{z\in \mathcal{Z}}\Pr\nolimits_{0}[Y=1|z]
\end{equation*}

For any value of $\theta =(\alpha ,\beta )$ the value of $\alpha $ defines
the sets $S_{j}$ above and the value of $\beta $ defines the bivariate
normal distribution to be employed. Four numerical integrations deliver $%
F_{U}(S_{j},\beta )$ for $j\in \{1,2,3,4\}$ from which probabilities for the
other regions are obtained by addition. Initially searching over a grid of
values of $\theta $ will produce the identified set. An optimisation
approach will be more effective.

%TCIMACRO{\TeXButton{B}{\begin{table}[tbp] \centering}}%
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\begin{table}[tbp] \centering%
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\begin{tabular}{|c|c|c|c|}
\hline
$j$ & $S_{j}$: unions of elemental sets & $x_{1}$ & $x_{2}$ \\ \hline\hline
\multicolumn{1}{|l|}{$1$} & \multicolumn{1}{|l|}{$a(0,0,h)$} & 
\multicolumn{1}{|l|}{} & \multicolumn{1}{|l|}{} \\ \hline
\multicolumn{1}{|l|}{$2$} & \multicolumn{1}{|l|}{$a(0,1,h)$} & 
\multicolumn{1}{|l|}{} & \multicolumn{1}{|l|}{} \\ \hline
\multicolumn{1}{|l|}{$3$} & \multicolumn{1}{|l|}{$a(1,0,h)$} & 
\multicolumn{1}{|l|}{} & \multicolumn{1}{|l|}{} \\ \hline
\multicolumn{1}{|l|}{$4$} & \multicolumn{1}{|l|}{$a(1,1,h)$} & 
\multicolumn{1}{|l|}{} & \multicolumn{1}{|l|}{} \\ \hline
\multicolumn{1}{|l|}{$5$} & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(0,1,h)$} & 
\multicolumn{1}{|l|}{$0$} & \multicolumn{1}{|l|}{} \\ \hline
\multicolumn{1}{|l|}{$6$} & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(1,0,h)$} & 
\multicolumn{1}{|l|}{} & \multicolumn{1}{|l|}{$0$} \\ \hline
\multicolumn{1}{|l|}{$7$} & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(1,1,h)$} & 
\multicolumn{1}{|l|}{} & \multicolumn{1}{|l|}{} \\ \hline
\multicolumn{1}{|l|}{$8$} & \multicolumn{1}{|l|}{$a(0,1,h)\cup a(1,0,h)$} & 
\multicolumn{1}{|l|}{} & \multicolumn{1}{|l|}{} \\ \hline
\multicolumn{1}{|l|}{$9$} & \multicolumn{1}{|l|}{$a(0,1,h)\cup a(1,1,h)$} & 
\multicolumn{1}{|l|}{} & \multicolumn{1}{|l|}{$1$} \\ \hline
\multicolumn{1}{|l|}{$10$} & \multicolumn{1}{|l|}{$a(1,0,h)\cup a(1,1,h)$} & 
\multicolumn{1}{|l|}{$1$} & \multicolumn{1}{|l|}{} \\ \hline
\multicolumn{1}{|l|}{$11$} & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(0,1,h)\cup
a(1,0,h)$} & \multicolumn{1}{|l|}{$0$} & \multicolumn{1}{|l|}{$0$} \\ \hline
\multicolumn{1}{|l|}{$12$} & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(0,1,h)\cup
a(1,1,h)$} & \multicolumn{1}{|l|}{$0$} & \multicolumn{1}{|l|}{$1$} \\ \hline
\multicolumn{1}{|l|}{$13$} & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(1,0,h)\cup
a(1,1,h)$} & \multicolumn{1}{|l|}{$1$} & \multicolumn{1}{|l|}{$0$} \\ \hline
\multicolumn{1}{|l|}{$14$} & \multicolumn{1}{|l|}{$a(0,1,h)\cup a(1,0,h)\cup
a(1,1,h)$} & \multicolumn{1}{|l|}{$1$} & \multicolumn{1}{|l|}{$1$} \\ \hline
\multicolumn{1}{|l|}{$15$} & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(0,1,h)\cup
a(1,0,h)\cup a(1,1,h)$} & \multicolumn{1}{|l|}{$0,1$} & \multicolumn{1}{|l|}{%
$0,1$} \\ \hline
\end{tabular}%
\caption{All possible unions of elemental sets of values of U and  values of
y for which sets T(y,x;h) lie entirely within a set for stated values of X}%
\label{unions1}%
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\end{table}%
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\bigskip

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\begin{tabular}{|l|c|c|}
\hline
$j$ & $S_{j}$ & $F_{U}(S_{j},\beta )$ must at least equal: \\ \hline\hline
$5$ & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(0,1,h)$} & \multicolumn{1}{|l|}{$%
\delta _{1}^{0}(z)\Pr_{0}[Y=0|x_{1},z]$} \\ \hline
$6$ & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(1,0,h)$} & \multicolumn{1}{|l|}{$%
\delta _{2}^{0}(z)\Pr_{0}[Y=0|x_{2},z]$} \\ \hline
$9$ & \multicolumn{1}{|l|}{$a(0,1,h)\cup a(1,1,h)$} & \multicolumn{1}{|l|}{$%
\delta _{2}^{0}(z)\Pr_{0}[Y=1|x_{2},z]$} \\ \hline
$10$ & \multicolumn{1}{|l|}{$a(1,0,h)\cup a(1,1,h)$} & \multicolumn{1}{|l|}{$%
\delta _{1}^{0}(z)\Pr_{0}[Y=1|x_{1},z]$} \\ \hline
$11$ & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(0,1,h)\cup a(1,0,h)$} & 
\multicolumn{1}{|l|}{$\delta _{1}^{0}(z)\Pr_{0}[Y=0|x_{1},z]+\delta
_{2}^{0}(z)\Pr_{0}[Y=0|x_{2},z]$} \\ \hline
$12$ & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(0,1,h)\cup a(1,1,h)$} & 
\multicolumn{1}{|l|}{$\delta _{1}^{0}(z)\Pr_{0}[Y=0|x_{1},z]+\delta
_{2}^{0}(z)\Pr_{0}[Y=1|x_{2},z]$} \\ \hline
$13$ & \multicolumn{1}{|l|}{$a(0,0,h)\cup a(1,0,h)\cup a(1,1,h)$} & 
\multicolumn{1}{|l|}{$\delta _{1}^{0}(z)\Pr_{0}[Y=1|x_{1},z]+\delta
_{2}^{0}(z)\Pr_{0}[Y=0|x_{2},z]$} \\ \hline
$14$ & \multicolumn{1}{|l|}{$a(0,1,h)\cup a(1,0,h)\cup a(1,1,h)$} & 
\multicolumn{1}{|l|}{$\delta _{1}^{0}(z)\Pr_{0}[Y=1|x_{1},z]+\delta
_{2}^{0}(z)\Pr_{0}[Y=1|x_{2},z]$} \\ \hline
\end{tabular}%
\caption{Unions of elemental sets of values of U delivering informative
bounds and bounding probabilities}\label{bounds1}%
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\end{table}%
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\subsection{$K$ valued $X$}

\textit{Just notes in this section at this point.}

\begin{itemize}
\item When $X$ is $K$-valued, with binary $Y$, there are $2^{K}$ elemental
sets $a(i_{1},\dots ,i_{K};h)$ of which at most $2K$ are non-empty.

\item Suppose the elements of $X$ are arranged in increasing or descending
(i.e. monotone) order, e.g.%
\begin{equation*}
x_{1}<x_{2}<\cdots <x_{K}.
\end{equation*}%
Then the potentially non-empty $a$ sets have entries in monotone order.

\begin{itemize}
\item Thus for $K=2$ all elemental sets are potentially non-empty
(reflecting the fact that $2^{2}=2\times 2$).

\item With $K=3$ there are $2\times 3=6$ potentially non-empty elemental
sets out of a possible $2^{3}=8$. The guaranteed empty sets are $a(0,1,0;h)$
and $a(1,0,1;h)$.

\item When $K=4$ there are $8$ potentially non-empty elemental sets out of a
possible $16$. The $8$ guaranteed empty sets are as follows. 
\begin{equation*}
a(0,1,0,0;h)\quad a(0,0,1,0;h)
\end{equation*}%
\begin{equation*}
a(1,0,0,1;h)\quad a(1,0,1,0;h)\quad a(0,1,1,0;h)\quad a(0,1,0,1;h)
\end{equation*}%
\begin{equation*}
a(1,1,0,1;h)\quad a(1,0,1,1;h)
\end{equation*}

\item It is straightforward to enumerate the potentially non-empty elemental
sets. Order the values of $X$ so that%
\begin{equation*}
x_{1}<x_{2}<\cdots <x_{K}.
\end{equation*}%
Moving clockwise around the point $(\alpha _{0},-\alpha _{1})$ the sets
arise as follows.%
\begin{multline*}
a(1,1,1,\dots ,1,1,1),\quad a(1,1,1,\dots ,1,1,0),\quad a(1,1,1,\dots
,1,0,0),\quad a(1,1,1,\dots ,0,0,0)\dots \\
\dots ,a(1,1,0,\dots ,0,0,0),\quad a(1,0,0,\dots ,0,0,0),\quad a(0,0,0,\dots
,0,0,0), \\
a(0,0,0,\dots ,0,0,1),\quad a(0,0,0,\dots ,0,1,1),\quad a(0,0,0,\dots
,1,1,1),\dots \\
\dots a(0,0,1,\dots ,1,1,1),\quad a(0,1,1,\dots ,1,1,1)
\end{multline*}%
We start with all $i_{1}=i_{2}=\cdots i_{K}=1$ and then replace $1$ by $0$
in turn from the right until we arrive at the pie slice opposite\ $%
a(1,1,1,\dots ,1,1,1)$ which is $a(0,0,0,\dots ,0,0,0)$. We then replace $0$
by $1$ again in turn from the right until we arrive at the pie slice $%
a(0,1,1,\dots ,1,1,1)$ which is immediately to the left of $a(1,1,1,\dots
,1,1,1)$.
\end{itemize}

\item As stated above, at most $2K$ elemental sets are non-empty. In
principle we have to consider all possible unions of these. There are $%
2^{2K} $ such unions. But there are many that do not have to be considered.

\begin{itemize}
\item No \emph{disconnected} unions need be considered. If inequalities hold
for connected components of such sets then they hold for the unions of such
components (there is no intersection).
\end{itemize}

\item Each observable event $T(y,x;h)$ event is the union of $K$ elemental
sets.

\begin{itemize}
\item No unions obtained by leaving out more than $K$ elemental sets need be
considered.

\item Considering unions obtained by leaving out $S\leq K$ elemental sets,
only unions obtained by leaving out connected unions of elemental sets need
be considered (otherwise the union would be disconnected).

\item For all $S\leq K$ there are $2K$ unions of elemental events obtained
by leaving out $S$ elemental events which constitute a connected set.

\item It follows that there are at most $2K^{2}$ inequalities to be
considered, i.e. generating lower bounds which may be non-zero. For $K\in
\{2,3,4,5\}$ this is $\{8,18,32,50\}$ inequalities.
\end{itemize}

\item Given any particular union we include 
\begin{equation*}
\delta _{j}(z)\Pr [Y=y|x_{j},z]
\end{equation*}%
in the sum of probabilities constituting the lower bound if and only if
there are \emph{precisely} $K$ entries equal to $y$ in the $j$th position in
the terms $a(i_{1},\dots ,i_{K};h)$ appearing in the union.
\end{itemize}

\section{Random sets}

Define set valued random variables as follows.%
\begin{equation*}
T(Y,x;h)\equiv \{u:h(x,u)=Y\}
\end{equation*}%
\begin{equation*}
T(Y,X;h)\equiv \{u:h(X,u)=Y\}
\end{equation*}%
Expressions given above are obtained through evaluating the containment
functional for these random sets using the set $S$, as follows.%
\begin{equation*}
\Pr\nolimits_{0}[T(Y,x;h)\subseteq S|x,z]=\sum_{i\in
\{0,1\}}1[T(i,x;h)\subseteq S]\Pr\nolimits_{0}[Y=i|X=x,Z=z]
\end{equation*}
\begin{equation*}
\Pr\nolimits_{0}[T(Y,X;h)\subseteq S|z]=\int \sum_{i\in
\{0,1\}}1[T(i,x;h)\subseteq S]\Pr\nolimits_{0}[Y=i|X=x,Z=z]dF_{X|Z=z}^{0}(x)
\end{equation*}%
In consequence the identified set defining inequalities given earlier can be
expressed succinctly as follows.%
\begin{equation}
F_{U}(S,\beta )\geq \max_{z\in \mathcal{Z}}\Pr\nolimits_{0}[T(Y,X;h)%
\subseteq S|z]
\end{equation}

This is the route to proving sharpness with Artstein etc.

\section{Some computations}

\subsection{Structures generating probability distributions}

We need to see what the force of the identifying restrictions is.

Let's generate probability distributions as follows.%
\begin{equation*}
Y=h(X,U)\equiv \left\{ 
\begin{array}{ccrcl}
0 & , & -\infty < & U_{1} & \leq a_{0}+(a_{1}+U_{2})X \\ 
1 & , & a_{0}+(a_{1}+U_{2})X< & U_{1} & \leq \infty%
\end{array}%
\right. \qquad U%
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\indt%
%EndExpansion
Z
\end{equation*}%
\begin{eqnarray*}
X &=&1[X^{\ast }>0] \\
X^{\ast } &=&\delta _{1}Z+\delta _{2}U_{1}+\delta _{3}U_{2}+\delta _{4}V
\end{eqnarray*}%
\begin{equation*}
\left[ 
\begin{array}{c}
U_{1} \\ 
U_{2} \\ 
V%
\end{array}%
\right] \sim N\left( \left( 
\begin{array}{c}
0 \\ 
0 \\ 
0%
\end{array}%
\right) ,\left( 
\begin{array}{ccc}
1 & b_{1} & 0 \\ 
b_{1} & b_{2}+b_{1}^{2} & 0 \\ 
0 & 0 & 1%
\end{array}%
\right) \right) .
\end{equation*}%
So the parameters that have to be set in the calculations are, for an
initial examination as follows.%
\begin{equation*}
a_{0}=0,\quad a_{1}=1
\end{equation*}%
\begin{equation*}
\delta \equiv \left[ 
\begin{array}{cccc}
\delta _{1} & \delta _{2} & \delta _{3} & \delta _{4}%
\end{array}%
\right] =\left[ 
\begin{array}{cccc}
2 & 1/(3)^{1/2} & 1/(3)^{1/2} & 1/(3)^{1/2}%
\end{array}%
\right]
\end{equation*}%
\begin{equation*}
b_{1}=0\qquad b_{2}=1
\end{equation*}%
\begin{equation*}
\mathcal{Z}=\{-1,-1/2,0,1/2,1\}
\end{equation*}

Maybe in some \textquotedblleft experiments\textquotedblright\ we can adjust
the strength of the instrument by changing $\delta _{1}$.

\subsection{The probability distribution $F_{YX|Z}^{0}$ with binary $X$}

First consider $\Pr\nolimits_{0}[X=0|Z=z]$. There is:%
\begin{equation*}
X=0\Leftrightarrow \left( \delta _{2}U_{1}+\delta _{3}U_{2}+\delta _{4}V\leq
-\delta _{1}Z\right)
\end{equation*}%
and%
\begin{equation*}
W\equiv \delta _{2}U_{1}+\delta _{3}U_{2}+\delta _{4}V\sim N(0,\lambda )
\end{equation*}%
where%
\begin{equation}
\lambda =\delta _{2}^{2}+2\delta _{2}\delta _{3}b_{1}+\delta
_{3}^{2}(b_{2}+b_{1}^{2})+\delta _{4}^{2}  \label{lambda_definition}
\end{equation}%
and so we have%
\begin{equation*}
\Pr\nolimits_{0}[X=0|Z=z]=\Phi \left( \frac{-\delta _{1}z}{\lambda ^{1/2}}%
\right)
\end{equation*}%
\begin{equation*}
\Pr\nolimits_{0}[X=1|Z=z]=1-\Phi \left( \frac{-\delta _{1}z}{\lambda ^{1/2}}%
\right) .
\end{equation*}

Now consider $\Pr\nolimits_{0}[Y=0\wedge X=0|Z=z]$. There is:%
\begin{equation*}
\left( Y=0\wedge X=0\right) \Leftrightarrow \left( \delta _{2}U_{1}+\delta
_{3}U_{2}+\delta _{4}V\leq -\delta _{1}z\right) \wedge (U_{1}\leq \alpha
_{0})
\end{equation*}%
and%
\begin{equation*}
\left[ 
\begin{array}{c}
W \\ 
U_{1}%
\end{array}%
\right] \sim N\left( \left[ 
\begin{array}{c}
0 \\ 
0%
\end{array}%
\right] ,\left[ 
\begin{array}{cc}
\lambda & \delta _{2}+\delta _{3}b_{1} \\ 
\delta _{2}+\delta _{3}b_{1} & 1%
\end{array}%
\right] \right)
\end{equation*}%
and so $\Pr\nolimits_{0}[Y=0\wedge X=0|Z=z]$ is the normal orthant
probability%
\begin{equation*}
\Pr [\left( W\leq -\delta _{1}z\right) \wedge \left( U_{1}\leq \alpha
_{0}\right) ].
\end{equation*}

Now consider $\Pr\nolimits_{0}[Y=0\wedge X=1|Z=z]$. There is:%
\begin{equation*}
\left( Y=0\wedge X=1\right) \Leftrightarrow (U_{1}\leq
a_{0}+a_{1}+U_{2})\wedge \left( \delta _{2}U_{1}+\delta _{3}U_{2}+\delta
_{4}V>-\delta _{1}z\right)
\end{equation*}%
and with $Q\equiv $ $U_{1}-U_{2}$ there is:%
\begin{equation*}
\left( Y=0\wedge X=1\right) \Leftrightarrow (Q\leq a_{0}+a_{1})\wedge \left(
W>-\delta _{1}z\right) .
\end{equation*}%
and%
\begin{equation*}
\left[ 
\begin{array}{c}
W \\ 
Q%
\end{array}%
\right] \sim N\left( \left[ 
\begin{array}{c}
0 \\ 
0%
\end{array}%
\right] ,\left[ 
\begin{array}{cc}
\lambda & (1-b_{1})(\delta _{2}+\delta _{3}b_{1})-\delta _{3}b_{2} \\ 
(1-b_{1})(\delta _{2}+\delta _{3}b_{1})-\delta _{3}b_{2} & 
(1-b_{1})^{2}+b_{2}%
\end{array}%
\right] \right)
\end{equation*}%
and $\Pr\nolimits_{0}[Y=0\wedge X=1|Z=z]$ is the normal orthant probability%
\begin{equation*}
\Pr [\left( W>-\delta _{1}z\right) \wedge \left( Q\leq a_{0}+a_{1}\right) ].
\end{equation*}%
Finally:%
\begin{equation*}
\Pr\nolimits_{0}[Y=1\wedge
X=0|Z=z]=\Pr\nolimits_{0}[X=0|Z=z]-\Pr\nolimits_{0}[Y=0\wedge X=0|Z=z]
\end{equation*}%
\begin{equation*}
\Pr\nolimits_{0}[Y=1\wedge
X=1|Z=z]=\Pr\nolimits_{0}[X=1|Z=z]-\Pr\nolimits_{0}[Y=0\wedge X=1|Z=z]
\end{equation*}

\subsection{$K$-valued $X$}

Now let $X$ take $K$ ordered values $x_{1}<x_{2}<\dots <x_{K}$. Let's
generate probability distributions as follows.%
\begin{equation*}
Y=h(X,U)\equiv \left\{ 
\begin{array}{ccrcl}
0 & , & -\infty < & U_{1} & \leq a_{0}+(a_{1}+U_{2})X \\ 
1 & , & a_{0}+(a_{1}+U_{2})X< & U_{1} & \leq \infty%
\end{array}%
\right. \qquad U%
%TCIMACRO{\TeXButton{indep}{\indt}}%
%BeginExpansion
\indt%
%EndExpansion
Z
\end{equation*}%
\begin{eqnarray*}
X &=&x_{k}\text{ iff }c_{k-1}<X^{\ast }\leq c_{k} \\
X^{\ast } &=&\delta _{1}Z+\delta _{2}U_{1}+\delta _{3}U_{2}+\delta _{4}V
\end{eqnarray*}%
\begin{equation*}
\left[ 
\begin{array}{c}
U_{1} \\ 
U_{2} \\ 
V%
\end{array}%
\right] \sim N\left( \left( 
\begin{array}{c}
0 \\ 
0 \\ 
0%
\end{array}%
\right) ,\left( 
\begin{array}{ccc}
1 & b_{1} & 0 \\ 
b_{1} & b_{2}+b_{1}^{2} & 0 \\ 
0 & 0 & 1%
\end{array}%
\right) \right) .
\end{equation*}

There is%
\begin{equation*}
\Pr\nolimits_{0}[X=x_{k}|Z=z]=\Phi \left( \frac{c_{k}-\delta _{1}z}{\lambda
^{1/2}}\right) -\Phi \left( \frac{c_{k-1}-\delta _{1}z}{\lambda ^{1/2}}%
\right) .
\end{equation*}%
with $\lambda $ defined as above in (\ref{lambda_definition}).

Now consider $\Pr\nolimits_{0}[Y=0\wedge X=x_{k}|Z=z]$. There is:%
\begin{equation*}
\left( Y=0\wedge X=x_{k}\right) \Leftrightarrow (U_{1}\leq
a_{0}+a_{1}x_{k}+x_{k}U_{2})\wedge \left( c_{k-1}-\delta _{1}z<W\leq
c_{k}-\delta _{1}z\right) 
\end{equation*}%
and with 
\begin{equation*}
Q_{k}\equiv U_{1}-x_{k}U_{2}
\end{equation*}%
there is:%
\begin{equation*}
\left( Y=0\wedge X=x_{k}\right) \Leftrightarrow (Q_{k}\leq
a_{0}+a_{1}x_{k})\wedge \left( c_{k-1}-\delta _{1}z<W\leq c_{k}-\delta
_{1}z\right) .
\end{equation*}%
and%
\begin{equation*}
\left[ 
\begin{array}{c}
W \\ 
Q_{k}%
\end{array}%
\right] \sim N\left( \left[ 
\begin{array}{c}
0 \\ 
0%
\end{array}%
\right] ,\left[ 
\begin{array}{cc}
\lambda  & \delta _{2}+b_{1}\left( \delta _{3}-\delta _{2}x_{k}\right)
-\delta _{3}x_{k}(b_{2}+b_{1}^{2}) \\ 
\delta _{2}+b_{1}\left( \delta _{3}-\delta _{2}x_{k}\right) -\delta
_{3}x_{k}(b_{2}+b_{1}^{2}) & (1-x_{k}b_{1})^{2}+x_{k}^{2}b_{2}%
\end{array}%
\right] \right) 
\end{equation*}%
and $\Pr\nolimits_{0}[Y=0\wedge X=x_{k}|Z=z]$ is%
\begin{equation*}
\Pr [\left( c_{k-1}-\delta _{1}z<W\leq c_{k}-\delta _{1}Z\right) \wedge
\left( Q_{k}\leq a_{0}+a_{1}x_{k}\right) ].
\end{equation*}%
which is easily calculated as the difference between two orthant
probabilities as follows.%
\begin{multline*}
\Pr\nolimits_{0}[Y=1\wedge X=x_{k}|Z=z]=\Pr [\left( W\leq c_{k}-\delta
_{1}Z\right) \wedge \left( Q_{k}\leq a_{0}+a_{1}x_{k}\right) ] \\
-\Pr [\left( W\leq c_{k-1}-\delta _{1}z\right) \wedge \left( Q_{k}\leq
a_{0}+a_{1}x_{k}\right) ].
\end{multline*}

Finally we have%
\begin{equation*}
\Pr\nolimits_{0}[Y=0\wedge
X=x_{k}|Z=z]=\Pr\nolimits_{0}[X=x_{k}|Z=z]-\Pr\nolimits_{0}[Y=0\wedge
X=x_{k}|Z=z]
\end{equation*}

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\end{document}
